# PLEASE HELP! supply me with answers and work1. in triangle ABC, a=3, b=5, c=7. What is m<C? 2. if tan A= (2/3) and sin B=(5/√41) and angles A and B are in Quadrant 1, find the value of...

PLEASE HELP! supply me with answers and work

1. in triangle ABC, a=3, b=5, c=7. What is m<C?

2. if tan A= (2/3) and sin B=(5/√41) and angles A and B are in Quadrant 1, find the value of tan(A+B).

3. if sin A=(2/3) where 0 degrees < A< 90 degrees, what is the value of sin2A?

4. In triangle ABC, m<A=32, a=12, and b=10. Find the measures of the missing angles and side of triangle ABC. Round each measure to the nearest tenth.

### 1 Answer | Add Yours

**1**. I will be using Heron's formula for finding the area of triangle, then use that to find the missing angle.

`A=sqrt(p(p-a)(p-b)(p-c))=sqrt(7.5*4.5*2.5*.5)~~6.50`

On the other hand `A=1/2*a*b*sinC => 6.50=1/2*3*5*sinC =>`

`sinC=13/15 =>m/_C=60`

**So angle C measures 60.**

**2.** We will be using the fact that `tan(A+B)=(tanA+tan B)/(1-tanA*TanB)`

Using the pythagorean theorem we can find that `tanB=5/4`

hence `tan(A+B)=(2/3+5/4)/(1-2/3*5/4)=`

`(8/12+15/12)/(1-10/12)=(8+15)/(12-10)=23/2`

So **tan(A+B)=11.5**

3.Here I will be using `sin2A=2sinAcosA`

Knowing that sine=opposite/hypotenuse, we can use the pythagorean theorem to find the adjacent side, hence the cosine.

`cosA=sqrt(9-4)/3=sqrt(5)/3`

so `sin2A=2*2/3*sqrt(5)/3=(4sqrt(5))/9`

**Thus measure of sin2A=`(4sqrt(5))/9` **